Polynomial rigidity of strong-field magnetic billiards

Abstract

A magnetic billiard describes a charged particle constrained to a planar domain: the particle moves along circular Larmor arcs in the interior and undergoes specular reflection at the boundary. The round disk has an explicit first integral that is polynomial in the velocity, and a central rigidity question asks whether any other smooth convex table can have such an integral. We answer this question negatively in the strong-field regime. Let $Ω\subset\mathbb R^2$ be a bounded strictly convex domain with smooth boundary $γ$, let $r=|B|^{-1}$ be the Larmor radius, and assume $0<r<r_0(γ)/2$, where $r_0(γ)$ is the maximal embedded tubular radius. If the magnetic billiard admits a nonconstant first integral polynomial in the velocity variables, of any finite degree, then $Ω$ is a disk. This removes the finite exceptional set of strong field strengths left by the earlier polynomial nonintegrability theory. The rigidity mechanism has two logically independent stages. First, the highest reflection mode gives a boundary winding identity. This determines the degree of the top coefficient and places all of its roots strictly inside the table, but it does not show that those roots coincide. Second, after the two leading reflection identities are continued to the normalization of the complexified boundary, their valuations at infinity exclude simultaneous poles of the coordinate functions. The remaining one-sided poles force the entire top coefficient to be one linear factor of multiplicity equal to the Fourier degree. Combining the location theorem with this root-collapse theorem produces a constant-angle relation between the boundary tangent and a radial direction, hence circularity. No real-analytic boundary hypothesis is imposed: analyticity follows from the algebraic strong-field parallel curves.